Question

Evaluate
$\sum _{n=1}^{\infty }{\tan }^{-1}\left(\frac{1}{n^2+n+1}\right)$

• A. $+\frac{\pi }{4}$
• B. $-\frac{\pi }{4}$
• C. $-\frac{\pi }{2}$
• D. $+\frac{\pi }{2}$

Answer 1

The correct option is A $+\frac{\pi }{4}$

Let $S=\sum _{n=1}^{\infty }{\tan }^{-1}\frac{1}{1+n+n^2}=\sum _{n=1}^{\infty }{\tan }^{-1}\frac{(n+1)-n}{1+n(n+1)}$

$=\sum _{n=1}^{\infty }{\tan }^{-1}(n+1)-{\tan }^{-1}n$

$=({\tan }^{-1}2-{\tan }^{-1}1)+({\tan }^{-1}3-{\tan }^{-1}2)+............+\left({\tan }^{-1}\right(n)-{\tan }^{-1}(n-1\left)\right)$

$={\tan }^{-1}n-{\tan }^{-1}1=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$ since $n\to \infty$